Problem Set 8
Problem 8.1. Method of Joints for Truss Analysis
Determine the force in each member of the truss, and state if the members
are in tension or compression.
Figure P8.1: Problem 8.1
Problem 8.2. Method of Sections for Truss Analysis
The roof truss supports the loading shown. Determine the force in the
members BC, CK, and KJ and state if these members are in tension or
compression.
Figure P8.2: Problem 8.2
Problem 8.3
Determine the force in each member of the truss and state if the members
are in tension or compression.
Solution
1. Mechanical System: truss.
2. Free-Body Diagram: joints C, D, and B.
P
P
3. Equations of equilibrium: Method of Joints Fx = 0 &
Fy = 0.
P1 = 4; P2 = 4 kN; a = 3 m; theta = 30 deg
% joint C
% Fy = 0 => FCB*sind(theta) - P2 = 0
FCB = P2/sind(theta) % FCB = 8 [kN] (T)
% Fx = 0 => FCD - FCB*cosd(theta) = 0
FCD = FCB*cosd(theta) % FCD = 6.9282 [kN] (C)
% joint D
% Fx = 0 => FDE - FCD = 0
FDE = FCD % FDE = 6.9282 [kN] (C)
% Fy = 0 => FDB - P1 = 0
FDB = P1 % FDB = 4 [kN] (T)
% joint B
% Fy’ = 0 => FBE*cosd(theta) - PDB*cosd(theta) = 0
FBE = FDB % FBE = 4 [kN] (C)
% Fx’ = 0 => (FDB+FBE)*sind(theta) + FCB - FBA = 0
FBA = (FDB+FBE)*sind(theta) + FCB % FBA = 12 [kN] (T)
Problem 8.4
The Howe bridge truss is subjected to the loading shown. Determine the
force in members HI, HB, and BC, and state if the members are in tension
or compression.
Figure P8.4: Problem 8.4
Solution
1. Mechanical System: bridge.
2. Free-Body Diagram: left-hand side of the bridge.
P
P
3. Equations of equilibrium: Method of Sections Fx = 0 &
Fy =
P
0 &
MP = 0, ∀P .
% support reactions
% sum M_E = 0 => 30*16+20*12+20*8+40*4-Ay*16 = 0
Ay = (30*16+20*12+20*8+40*4)/16 % Ay = 65 [kN]
% Fx = 0 =>
Ax = 0
% Method of Sections
% sum M_H = 0 => FBC*4+20*4+30*8-Ay*8 = 0
FBC = -(20*4+30*8-Ay*8)/4 % FBC = 50 [kN] (T)
% sum M_B = 0 => FHI*4+30*4-Ay*4 = 0
FHI = -(30*4-Ay*4)/4 % FHI = 35 [kN] (C)
% Fy = 0 => Ay-30-20-FHB*sind(45) = 0
FHB = (Ay-30-20)/sind(45) % FHB = 21.2 [kN] (C)
Problem 8.5
Determine the force in each member of the space truss and state if the
members are in tension or compression. The truss is supported by a balland-socket joints at A, B, and E.
Figure P 8.5: Problem 8.5
Solution
1. Mechanical System: space truss
2. Free-Body Diagram: nodes D and C.
P
P
3. Equations of equilibrium: Method of Joints Fx = 0 &
Fy =
P
0 &
Fz = 0.
Fx = -200; Fy = 400; % N
a = 2; b = 1.5; c = 5; d = 1; e = 2; % m
F = [Fx, Fy, 0];
rA = [0, 0, 0]; rB = [c+d, a+b, 0]; rC = [0, a+b, 0]; rD = [d, a, e];
uDA = -rD/norm(rD); % uDA = [ -0.333, -0.667, -0.667]
uDB = (rB-rD)/norm(rB-rD); % uDB = [ 0.894, 0.268, -0.358]
uDC = (rC-rD)/norm(rC-rD); % uDC = [ -0.371, 0.557, -0.743]
F_DA = FAD*uDA; F_DB = FBD*uDB; F_DC = FCD*uDC;
% sum F at node D
FD = F+F_DA+F_DB+F_DC;
solD = solve(FD(1),FD(2),FD(3));
% FAD = 343 [N] (T)
% FBD = 186 [N] (T)
% FCD = -397 [N] (C)
% node C
F_CD = -fCD*uDC
uCA = -rC/norm(rC); % uCA = [ 0, -1.0, 0]
uCB = (rB-rC)/norm(rB-rC); % uCB = [ 1.0, 0, 0]
uEC = [0, 0, 1];
F_AC = FAC*uCA; F_BC = FBC*uCB; F_EC = FEC*uEC;
FC = F_AC+F_BC+F_EC+F_CD;
solC = solve(FC(1),FC(2),FC(3));
% FAC = 221 [N] (T)
% FBC = 148 [N] (T)
% FEC = 295 [N] (C)